Find the unit digit in `7 ^(65) xx 6 ^(41) xx 3 ^(57).`

Options

A. 6
B. 7
C. 5
D. 8

Answer

To find the unit digit of the expression \( 7^{65} \times 6^{41} \times 3^{57} \), we can analyze each term separately and then combine the results.

### Step 1: Find the unit digit of \( 7^{65} \)

The unit digits of the powers of 7 repeat every 4 terms:
- \( 7^1 = 7 \) (unit digit 7)
- \( 7^2 = 49 \) (unit digit 9)
- \( 7^3 = 343 \) (unit digit 3)
- \( 7^4 = 2401 \) (unit digit 1)

This cycle (7, 9, 3, 1) repeats every 4 powers. To find the unit digit of \( 7^{65} \), we calculate \( 65 \mod 4 \):
\[ 65 \div 4 = 16 \quad \text{remainder } 1 \]
So, \( 65 \mod 4 = 1 \). This means the unit digit of \( 7^{65} \) is the same as the unit digit of \( 7^1 \), which is 7.

### Step 2: Find the unit digit of \( 6^{41} \)

The unit digit of any power of 6 is always 6:
- \( 6^1 = 6 \)
- \( 6^2 = 36 \)
- \( 6^3 = 216 \)
- \( 6^4 = 1296 \)

Thus, the unit digit of \( 6^{41} \) is 6.

### Step 3: Find the unit digit of \( 3^{57} \)

The unit digits of the powers of 3 also repeat every 4 terms:
- \( 3^1 = 3 \) (unit digit 3)
- \( 3^2 = 9 \) (unit digit 9)
- \( 3^3 = 27 \) (unit digit 7)
- \( 3^4 = 81 \) (unit digit 1)

This cycle (3, 9, 7, 1) repeats every 4 powers. To find the unit digit of \( 3^{57} \), we calculate \( 57 \mod 4 \):
\[ 57 \div 4 = 14 \quad \text{remainder } 1 \]
So, \( 57 \mod 4 = 1 \). This means the unit digit of \( 3^{57} \) is the same as the unit digit of \( 3^1 \), which is 3.

### Step 4: Combine the unit digits

Now we have:
- Unit digit of \( 7^{65} \) is 7.
- Unit digit of \( 6^{41} \) is 6.
- Unit digit of \( 3^{57} \) is 3.

Next, we multiply these unit digits together:
\[ 7 \times 6 = 42 \quad (\text{unit digit is } 2) \]
Now, we take this result and multiply it by the unit digit of \( 3^{57} \):
\[ 2 \times 3 = 6 \]

### Final Answer

Thus, the unit digit of \( 7^{65} \times 6^{41} \times 3^{57} \) is **6**.

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